LEAD ACID BATTERY
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From: budgie on 3 Apr 2006 20:33 On Mon, 03 Apr 2006 15:46:21 GMT, Robert Baer <robertbaer(a)earthlink.net> wrote: >budgie wrote: > >> On Sat, 01 Apr 2006 08:44:01 GMT, Robert Baer <robertbaer(a)earthlink.net> wrote: >> >> >>>budgie wrote: >>> >>> >>>>On Sat, 01 Apr 2006 08:47:10 +0800, budgie <me(a)privacy.net> wrote: >>>> >>>> >>>> >>>>>and as the O/P's required current is 700mA the redcution in performance would be >>>>>absolutely marginal, giving his target of 12 hours almost spot-on. >>>> >>>> >>>> >>>>Aaarrrggghh! My turn to stuff up! Ignore that line, >>>> >>>>Particularly if the O/P really meant: >>>> >>>>" I have to design a voltage source ( +/- 18volts, 13A )" >> >> >>> So? Use of three 6V batteries in series gives 18V. >> >> >> You totally missed the point. 13A? or 700mA? > 13A i believe was the batttery rating, and 700mA was the maximum >proposed load.. Yes, I also assumed 13Ah for the proposed batteries and 700mA for the load. But the O/P's statement "I have to design a voltage source ( +/- 18volts, 13A )" is at odds with that.
From: budgie on 3 Apr 2006 20:36 On 3 Apr 2006 11:29:12 -0700, "RHRRC" <h.lewis(a)connect-2.co.uk> wrote: > >budgie wrote: >> On 2 Apr 2006 08:13:20 -0700, "RHRRC" <h.lewis(a)connect-2.co.uk> wrote: >> >> > >> >budgie wrote: > ><snipped> > >> >If you have another methodology that you think is advantageousnby all >> >means advocate its use please do not use the symbol 'C' as this is >> >already used by the battery fraternity. >> > >> >How do you intend to measure battery capacity? >> >> I was commenting on the pedantry in your post, seemingly objecting to 'C' vs >> 'C-rate'. I do understand 'C' and its purpose/role TUVM. >> >> >> >> Review your other line: "The nominal 'C-rate' of a 13Ah battery is 13amps." >> >> >> >> Try reading the two in conjunction. >> >> Apaprently you haven't. > >I am glad that you understand what 'C' represets. Me too. >Using some data to hand a discharge of a typical 13Ah sla battery at >its nominal C-rate would be a disharge at 13A. Please post the source of this 'C-rate' shite you keep on about. >However a discharge of a typical 13Ah sla battery at the C rate would >be a disharge at 7.8 amps (Yuasa NP series data) So much for your "worldwide standard". >Sorry to be so pedantic as to point out the difference between 13 and >7.8 - in your world this is a trivial difference but for the rest it is >more than likely significant.
From: Rich Grise on 4 Apr 2006 17:38 On Mon, 03 Apr 2006 13:26:53 -0500, John Fields wrote: > On Mon, 03 Apr 2006 15:46:21 GMT, Robert Baer <robertbaer(a)earthlink.net> >>budgie wrote: > >>> You totally missed the point. 13A? or 700mA? >> 13A i believe was the batttery rating, and 700mA was the maximum >>proposed load.. > > 13A has nothing to do with it. > > The _capacity_ of the battery is 13AH (ampere-hours) at a 20hour rate, > which means the voltage at the initially fully charged battery's terminals > will fall to 5.25V when a constant current of > > 13AH > ------ = 0.65A > 20H > > is drawn from the battery, continuously, for a period of 20 hours. Yes, that's true, and the OP seems to have said that his app called for 700 mA, so he'd get 650 ----- * 20H = do I really have to do the math? ;-) 700 Cheers! Rich
From: John Fields on 4 Apr 2006 18:20 On Tue, 04 Apr 2006 21:38:19 GMT, Rich Grise <richgrise(a)example.net> wrote: >On Mon, 03 Apr 2006 13:26:53 -0500, John Fields wrote: >> On Mon, 03 Apr 2006 15:46:21 GMT, Robert Baer <robertbaer(a)earthlink.net> >>>budgie wrote: >> >>>> You totally missed the point. 13A? or 700mA? >>> 13A i believe was the batttery rating, and 700mA was the maximum >>>proposed load.. >> >> 13A has nothing to do with it. >> >> The _capacity_ of the battery is 13AH (ampere-hours) at a 20hour rate, >> which means the voltage at the initially fully charged battery's terminals >> will fall to 5.25V when a constant current of >> >> 13AH >> ------ = 0.65A >> 20H >> >> is drawn from the battery, continuously, for a period of 20 hours. > >Yes, that's true, and the OP seems to have said that his app called for >700 mA, so he'd get > > 650 > ----- * 20H = do I really have to do the math? ;-) > 700 --- Well, yes... And then you need to backtrack through the thread to find out what the OP _really_ wanted. Sound like fun? -- John Fields Professional Circuit Designer
From: Rich Grise on 6 Apr 2006 14:40
On Tue, 04 Apr 2006 17:20:49 -0500, John Fields wrote: .... > And then you need to backtrack through the thread to find out what > the OP _really_ wanted. Sound like fun? Nothing to it: ------------------------------------ On Wed, 29 Mar 2006 10:49:27 -0800, john wrote: Hi, I have to design a voltage source ( +/- 18volts, 13A ) using lead acid batteries. My circuit draw is 700mA and requires plus minus 18 volts to operate efficiently. I need batteries that can atleast run for 10 to 12 hours before the voltage drops to +/- 17 volts. I am thinking of adding three 6 volts, 13AH ( rated for 20AH ) batteries in series to produce +18 volts and adding three 6 volts to generate -18 volts. I choose the battery ( BP13-6V ), http://www.zbattery.com/zbattery/ub13-6.html. Can anybody advice me that am I doing the right thing that will these six batteries last for 10 to 12 hours maintaining +/- 18 volts @ 600mA. Thanks Regards John ------------------------------------ I guess when I saw "My circuit draw is 700 mA" I assumed that he means that his circuit draws 700 mA. I also assume that when he said "18 volts, 13A", he really meant the 13 AH capacity that he mentions five lines later, and the lack of the "H" in the first line is an oversignt. The batteries he linked to are, in fact, 13 AH. Hope This Helps! Rich |